The rate of radiation by a black body is R at | vedclass

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(C) The rate of radiation for a black body is given by the Stefan-Boltzmann law: $R = \left(\frac{dQ}{dt}\right)_1 = e A \sigma T^4$. Since it is a bl

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$16.2$

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(C) The rate of radiation for a black body is given by the Stefan-Boltzmann law: $R = \left(\frac{dQ}{dt}ight)_1 = e A \sigma T^4$. Since it is a black body,emissivity $e = 1$,so $R = A \sigma T^4$. For the second body,the rate of radiation is: $\left(\frac{dQ}{dt}ight)_2 = e' A' \sigma (T')^4$. Given $e' = 0.2$,$A' = A$,and $T' = 3T$: $\left(\frac{dQ}{dt}ight)_2 = 0.2 	imes A 	imes \sigma 	imes (3T)^4$. $\left(\frac{dQ}{dt}ight)_2 = 0.2 	imes A 	imes \sigma 	imes 81 T^4$. $\left(\frac{dQ}{dt}ight)_2 = 16.2 	imes A \sigma T^4$. Substituting $R = A \sigma T^4$,we get: $\left(\frac{dQ}{dt}ight)_2 = 16.2 R$.

The rate of radiation by a black body is $R$ at temperature $T$. Another body has the same area but an emissivity of $0.2$ and a temperature of $3T$. Its rate of radiation is: (in $R$)

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